What is more efficient, i++ or ++i? [duplicate]

问题

This question already has answers here:

Closed 10 years ago.

Exact Duplicate: Is there a performance difference between i++ and ++i in C++?
Exact Duplicate: Difference between i++ and ++i in a loop?

---

What is more efficient, i++ or ++i?

I have only used this in Java and C/C++, but I am really asking for all languages that this is implemented in.

In college I had a professor show us that ++i was more efficient, but it has been a couple of years, and I would like to get input from the Stack Overflow community.

回答1:

i++ :

• create a temporary copy of i
• increment i
• return the temporary copy

++i :

• increment i
• return i

With optimizations on, it is quite possible that the resulting assembly is identical, however ++i is more efficient.

edit : keep in mind that in C++, i may be whatever object that support the prefix and postfix ++ operator. For complex objects, the temporary copy cost is non negligible.

回答2:

I would look elsewhere for optimization potential.

回答3:

Efficiency shouldn't be your concern: it is meaning. The two are not the same, unless they are freestanding: one operates pre-use of the value, the other post.

int i; i = 1; cout << i++; //Returns 1

int i; i = 1; cout << ++i; //Returns 2

When meaning isn't important, most compilers will translate both ++i and i++ (say in a for loop) into the same machine/VM code.

回答4:

It does not matter on a modern compiler.

int v = i++;  

is the same as

int v = i;
i = i + 1;

A modern compiler will discover that v is unused and the code to calculate v is pure (no side effects). Then it will remove v and the assignment code and will generate this

i = i + 1;

回答5:

It does matter! Especially if you're on C++...

++i // the prefered way, unless..
auto j = i++ // this is what you need

You should always use the prefix notation to avoid necessary copying overhead and with C++ this matters!

回答6:

++i is potentially more efficient for a non-trivial implementation of operator++, but even in that scenario, the compiler may be able to optimize away the intermediate temporary.

回答7:

++i doesn't need a temporary variable to store stuff in. Think of them like this:

++i

int preIncrement(int i)
{
    i = i + 1;
    return i;
}

i++

int i = 5; // as an example
int postIncrement(_i)
{
    int temp = _i;
    i = _i + 1;
    return temp;
}

See? Postincrement requires a temporary variable. Assuming the compiler doesn't sort it all out for you, which it almost certainly does.

Of course, more important is program logic; you run the risk of encountering The Sad Tragedy of Micro-Optimisation Theatre if you worry about this too much...:)

回答8:

Well, in C++ I believe they have different uses, depending on when you want the variable updated.

Efficiency shouldn't determine when you use one over the other, but I would assume they would have the same efficiency either way.

回答9:

Unless I'm missing something, they should have the same efficiency. They should both result in a single add instruction. It's just a matter of where the add instruction takes place: at the beginning or the end of your line of code.

回答10:

++i is faster because i++ has to store i, then increment it, then return the stored value of i. ++i simply increments i then returns it.

// ++i
i += 1;
return i;

// i++
temp = i;
i += 1;
return temp;

回答11:

A stand-alone "i++;" or "++i;" should generate equally efficient code. The difference comes if you're using it in an expression, where the "side effect" comes into play.

That said, there was a time, back when "all the world's a Vax", and compilers sucked, that ++i was said to be more efficient that i++, even in a "for (i = 0; i < N; ++i)" type setting.

回答12:

In general, it is more efficient to use ++i than i++. The simple reason for this is that ++i is utterly the same as

i += 1;

which for x86 is a single instruction (and likely most other widely used architectures). i++ is however equal to

tmp = i; i += 1;

That is because the old value of 'i' is what i++ evaluates to. And clearly that requires more work than simply i += 1;

But as stated above, this has virtually no impact with a sufficiently clever compiler, as it will optimize unused operations away. For many interpreted languages (example: PHP) there is likely a minimal gain in speed for the ++i; But this increase is negligible.

回答13:

It's generally easier to type i++, hence it is more efficient in terms of productivity time.

Seriously, though, if i is a native data type (such as int, double, etc) -- no difference.

And it is implementation depended if it's a user-defined type such as

class Type
{
    Type& operator ++(){}
    const Type& operator ++(int i){}
};  

T i;

回答14:

++i takes one less processor instruction than i++ in x86 assembly without optimization.

回答15:

There is no difference. Use the construct that makes the most sense.

If your application runs slowly, I can guarantee you that it will never be because of speed differences in the integer increment operation. If it is, it's a severe bug in the compiler. Speed problems in your application will be algorithmic inefficiencies, waiting for I/O, and so on.

Don't worry about problems that you don't have. Premature optimization is the root of all evil.

回答16:

This Stack Overflow question has a great answer: Is there a performance difference between i++ and ++i in C?

I would like to add that you should use whichever one suits your needs better. Except in the most time critical of applications, it is not important. From an academic perspective too, it is better to write code that expresses what you need and optimize at last.

回答17:

There is no right or wrong answer.

As it depends on:

1. How it was implemented by the compiler.

2. What CPU the system runs on.

3. If i is byte or i is double word

回答18:

It depends upon the context, for example:

x = i++

In this case, 'x' will be equal to 'i', and only after that will 'i' be increased by one.

x = ++i

In this case, 'i' will be increase by one and then the new value of 'x' will be assigned to 'x'.

In the case of a 'for' loop, there is little apparent difference other than performance (++i is faster).

回答19:

Generally, in C++, postfix will require the additional construction of the object incremented, while prefix is applied directly to the object. (Or so I've read.)

As I am unable to attest to how the compiler handles it due to my limited knowledge on the matter, it could be handled for you making it a moot point.

回答20:

It's hard to answer this precisely as it depends on the compiler/interpreter implementation.

But generally speaking you can say roughly extend i++ to the following instructions:

COPY i to tmp
INCREMENT tmp
SAVE tmp as i

While ++i will roughly extend to:

LOAD i
INCREMENT i

You can't just say that ++i is faster than i++ since language implementations are pretty smart and they can optimize these instructions when you know that you won't access the temporary value of i++. This usually happens in say a for loop. So in many cases it's just the same.

If you're trying to these kind of micro-optimizations I'd advice you to profile/measure before chosing one over another.

来源：https://stackoverflow.com/questions/561588/what-is-more-efficient-i-or-i